Parameter transfer to functions via pointer in C

Question

Hi,

i got a code from a book of my university and it does not work. Error message:

warning: 'swap' defined but not used [-Wunused-function]
void swap(int *a, int *b)
^~~~
..\Playground\~1\AppData\Local\Temp\cc2dRKGB.o:source.c:(.text+0x46): undefined reference to `swap'
collect2.exe: error: ld returned 1 exit status

Even if i try other online C Compilers the message is the same.
(I simply want to change two Variables)

Code:
#include <stdio.h>
void swap(int *a, int *b);
int main()
{
int zahl1, zahl2;
zahl1 = 10;
zahl2 = 20;
printf("Startvalue: Zahl1: %i, Zahl2: %i", zahl1, zahl2);
swap(&zahl1, &zahl2);
printf("Finalvalue: Zahl1: %i, Zahl2: %i", zahl1, zahl2);
return 0;
void swap(int *a, int *b)
{
int vhilf;
vhilf = *a;
*a = *b;
*b = vhilf;
}
}

Answer 1

Your swap() function needs to be defined outside of main() and at least its prototype needs to be provided before the main() function.

Answer 2

as Sonic have said, just erase the last semicolon and put one after the return 0